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integrate frm -1 to 1: [x+[x+[x]]]dx
Don't you know that you can cut this integration to two parts?
ie, -1∫1 f(x).dx = -1∫0 f(x).dx + 0∫1 f(x).dx
Let f(x) = [x]
By the defenition of [x]; we have,
f(x) = x, x>0 and f(x) = -x, x<0
Then, come to your problem,
Let g(x) = [x+[x+[x]]].
When x>0; g(x) = x+x+x = 3x;
When x<0; g(x) = -x; (Why?? Do it yourself!!)
Then your problem reduces to:
-1∫1 [x+[x+[x]]].dx
= -1∫0 (-x).dx + 0∫1 3x.dx
= -(0/2) - (-1/2) + (3*1/2) - 0
= +(1/2) + (3/2)
= 2.
Is this the correct answer??
Sorry Sorry! I misunderstood your question! I'm extremely sorry!
I thought that [x] is modulus function! I thought of "greatest integer function" just after clicking "POST REPLY" button!
I couldn't undo it!
Again, extremely sorry for wasting your time!!
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